Spectrum questions almost never ask you to calculate. They hand you a wavelength, or a band name, and ask where it sits. The mark is lost by sliding one rung up or down the ladder — and the corpus records exactly one recurring slip for this pattern: swapping microwave for infrared.
Fix the ladder once, ordered by wavelength, longest first:
Frequency runs the opposite way. Longest wavelength means lowest frequency, so radio sits at the bottom of the frequency ladder and gamma at the top. Half the band-ordering questions in this unit are just that inversion, asked once in wavelength and once in frequency.
Two details from NCERT Class 12 Physics Chapter 8, page 208, decide the harder questions. First, the quoted ranges overlap — ultraviolet runs down to roughly 6 × 10⁻¹⁰ m while X-rays begin near 1 × 10⁻⁹ m. There is no sharp division anywhere in the spectrum. A wave in the overlap is classified by how it was produced, not by a boundary number. Second, the spectrum is continuous: it is one unbroken range of wavelengths that has been carved into named bands for convenience.
For NEET this topic is low-yield but low-cost: roughly one question every two or three papers, recall-weighted, answerable in thirty seconds if the ladder is memorised and in zero seconds if it is not.
Watch-out: when a wavelength is quoted in millimetres or nanometres, convert to metres before comparing with a boundary. "0.6 mm" reads as small, and small reads as microwave to a tired eye — but 6 × 10⁻⁴ m is shorter than 1 × 10⁻³ m, which puts it in infrared.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which band lies immediately beyond the violet end of the visible region, on the shorter-wavelength side?
Answer: C. Ultraviolet begins where visible light ends at about 4 × 10⁻⁷ m and extends to shorter wavelengths, as listed in the spectrum section of NCERT Class 12 Physics Chapter 8, page 208.
Why A is wrong: A is wrong because infrared lies immediately beyond the RED end of the visible region, at longer wavelength than visible light, not beyond violet.
Why B is wrong: B is wrong because microwaves are two rungs further out on the long-wavelength side, lying between radio waves and infrared.
Why D is wrong: D is wrong because radio waves occupy the longest-wavelength end of the whole spectrum, the far side from violet.
X-rays are commonly produced by which of the following processes?
Answer: A. The standard laboratory source of X-rays is the sudden deceleration of high-energy electrons striking a metal target, given as the production method for that band in NCERT Class 12 Physics Chapter 8, page 208.
Why B is wrong: B is wrong because nuclear transitions in an excited nucleus produce gamma rays, the band lying beyond X-rays at shorter wavelength.
Why C is wrong: C is wrong because accelerating charges in an aerial driven by an oscillating circuit produce radio waves, at the opposite end of the spectrum.
Why D is wrong: D is wrong because klystrons and magnetrons are the special vacuum tubes used to generate microwaves.
Among ultraviolet, X-rays, infrared and microwaves, which band has the longest wavelength?
Answer: D. Microwaves span 0.1 m down to 1 × 10⁻³ m, longer than every other band in the list, per the spectrum ordering in NCERT Class 12 Physics Chapter 8, page 208.
Why A is wrong: A is wrong because ultraviolet sits on the short-wavelength side of visible light, below 4 × 10⁻⁷ m.
Why B is wrong: B is wrong because X-rays have the shortest wavelength in this list, around 1 × 10⁻⁹ m and below.
Why C is wrong: C is wrong because infrared, although longer than visible, stops at 1 × 10⁻³ m where microwaves begin; picking infrared here is the microwave-for-infrared swap.
A monochromatic radiation has a wavelength of 5.5 × 10⁻⁷ m in vacuum. It belongs to which band?
Answer: B. The visible band runs from 4 × 10⁻⁷ m to 7 × 10⁻⁷ m, and 5.5 × 10⁻⁷ m lies inside it, as tabulated in NCERT Class 12 Physics Chapter 8, page 208.
Why A is wrong: A is wrong because the ultraviolet band lies below 4 × 10⁻⁷ m, and 5.5 × 10⁻⁷ m is longer than that boundary.
Why C is wrong: C is wrong because infrared begins above 7 × 10⁻⁷ m, and 5.5 × 10⁻⁷ m is shorter than that boundary.
Why D is wrong: D is wrong because microwaves lie between 1 × 10⁻³ m and 0.1 m, more than three orders of magnitude longer than the given wavelength.
Radiation of wavelength 3.0 × 10⁻⁵ m is incident on a detector. The radiation is:
Answer: D. Infrared occupies 1 × 10⁻³ m down to 7 × 10⁻⁷ m, and 3.0 × 10⁻⁵ m sits comfortably inside that span, per NCERT Class 12 Physics Chapter 8, page 208.
Why A is wrong: A is wrong because the microwave band stops at 1 × 10⁻³ m, and 3.0 × 10⁻⁵ m is about thirty times shorter than that boundary; this is the swap the pattern's distractor exploits.
Why B is wrong: B is wrong because visible light runs only from 4 × 10⁻⁷ m to 7 × 10⁻⁷ m, far shorter than the given wavelength.
Why C is wrong: C is wrong because ultraviolet is shorter still, lying below 4 × 10⁻⁷ m.
Arrange infrared, ultraviolet, microwave and visible radiation in order of increasing frequency.
Answer: A. Frequency rises as wavelength falls, and the wavelength order is microwave > infrared > visible > ultraviolet, so the frequency order is the reverse — matching the band ordering in NCERT Class 12 Physics Chapter 8, page 208.
Why B is wrong: B is wrong because it swaps microwave and infrared: microwave has the longer wavelength, so it has the lower frequency and must come first.
Why C is wrong: C is wrong because that is the order of increasing wavelength, which is exactly the reverse of increasing frequency.
Why D is wrong: D is wrong because it places visible below infrared; visible light has the shorter wavelength of the two and therefore the higher frequency.
A source emits two spectral lines of wavelengths λ₁ = 4.0 × 10⁻⁸ m and λ₂ = 4.0 × 10⁻¹¹ m in vacuum. Identify the bands and state which line has the higher frequency.
Answer: C. λ₁ = 40 nm falls inside the ultraviolet span (4 × 10⁻⁷ m down to about 6 × 10⁻¹⁰ m) and λ₂ = 0.04 nm inside the X-ray span; the shorter wavelength always carries the higher frequency, per NCERT Class 12 Physics Chapter 8, page 208.
Why A is wrong: A is wrong on the second step only: the two bands are named correctly, but the shorter wavelength λ₂ carries the higher frequency, not λ₁.
Why B is wrong: B is wrong because 4.0 × 10⁻⁸ m is 40 nm, inside ultraviolet; the X-ray band begins only at about 1 × 10⁻⁸ m (10 nm).
Why D is wrong: D is wrong because visible light stops at 4 × 10⁻⁷ m, and λ₁ is ten times shorter than that boundary while λ₂ lies far beyond ultraviolet.
NCERT quotes the ultraviolet band as extending down to about 6 × 10⁻¹⁰ m while the X-ray band is quoted as beginning near 1 × 10⁻⁸ m, so the two stated ranges overlap. The correct reading of this overlap is that:
Answer: B. NCERT Class 12 Physics Chapter 8, page 208, states that the spectrum is continuous with no sharp division between one part and the next, and that overlapping bands are distinguished by their method of production.
Why A is wrong: A is wrong because the overlap is stated deliberately rather than by mistake; the spectrum is continuous and its named divisions are a convenience, not a physical partition.
Why C is wrong: C is wrong because radiation in the overlap region is perfectly real and is named by its production mechanism, not discarded as unclassifiable.
Why D is wrong: D is wrong because all electromagnetic waves travel at the same speed in vacuum; speed plays no part in how the bands are divided.
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Given.
λ = 6.0 × 10⁻⁴ m (two significant figures), in vacuum.
Required.
The band name.
Concept.
The spectrum is one continuous range of wavelengths divided into named bands. Placing a wavelength means finding which pair of conventional boundary values it falls between. The two boundaries that matter here are 1 × 10⁻³ m (microwave on the long side, infrared on the short side) and 7 × 10⁻⁷ m (infrared on the long side, visible on the short side).
Formula.
None. This pattern carries no formula in the dossier and has calculation_load: none; the operative relation is the ordered wavelength ladder itself.
Substitution.
Compare λ with each boundary: is 6.0 × 10⁻⁴ m greater or less than 1.0 × 10⁻³ m, and greater or less than 7 × 10⁻⁷ m?
Calculation.
6.0 × 10⁻⁴ m = 0.60 × 10⁻³ m, which is less than 1.0 × 10⁻³ m. It is also clearly greater than 7 × 10⁻⁷ m. So λ falls between the two infrared boundaries. Note that the band-edge values 1 × 10⁻³ m and 7 × 10⁻⁷ m are conventional markers of classification, not measured quantities, so they place no limit on the significant figures of the given wavelength.
Final answer.
Infrared, near its long-wavelength edge — just inside the boundary with microwaves.
Common trap.
Reading "0.6 mm" as "under a millimetre, therefore microwave." The direction is what decides it: wavelengths shorter than 1 mm are infrared, longer than 1 mm are microwave. Convert to metres in scientific notation before comparing, and the direction stops being guesswork. This is the swaps-bands distractor the corpus records for this pattern.
Similar NEET-style question.
A transmitter emits radiation of wavelength 8.0 × 10⁻² m in vacuum. Name its band. *(Answer: 8.0 × 10⁻² m lies between 1 × 10⁻³ m and 0.1 m — microwave, near its long-wavelength edge.)*
Wavelength bands (decreasing): radio (>1 m) > microwave (1mm-1m) > infrared (700nm-1mm) > visible (400-700 nm) > UV (10-400 nm) > X-rays (0.01-10 nm) > gamma (<0.01 nm). All travel at c in vacuum.
-- NCERT Class 12 Physics, Ch. 8, p. 208More in Electromagnetic Waves: 2 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
swaps bands
Confuses microwave with infrared
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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